- $\abs{x} = 0$ if and only if $x = 0$.
- $\abs{xy} = \abs{x}\cdot \abs{y}$.
- There exists a real number $c$ such that for all $x,y\in K$, $\abs{x + y}\le c\max(\abs{x}, \abs{y})$.
Example Let $K = \BQ$. The function $\abs{\cdot}$ given by embedding $\BQ$ into $\BR$ and then taking absolute value is a valuation (henceforth we will denote this $\abs{\cdot}_\infty$. This notation is a little funky, but should make a more sense after seeing Ostrowski's theorem below).
Example / Exercise Let $K = \BQ$ and let $p$ be a rational prime. Let $v_p(x)$ be the exponent of the power of $p$ dividing $x$. Then the function given by $\abs{x} = p^{-v_p(x)}$ is a valuation on $\BQ$. (This is the standard $p$-adic valuation on $\BQ$, and is denoted $\abs{\cdot}_p$.)
Definition / Example The trivial valuation $\abs{\cdot}_0$ is the one that takes on the value of 1 on all of $K^\times$.
Proposition Suppose $\abs{\cdot}$ is a valuation on $K$, and let $r$ positive real number. Then $\abs{\cdot}^r$ is also a valuation on $K$.
Proof Obvious. $\boxed{}$
Definition Two nontrivial valuations $\abs{\cdot}_1$ and $\abs{\cdot}_2$ are equivalent if there exists a real number $r > 0$ such that $\abs{x}_1^r = \abs{x}_2$ for all $x\in K$.
Proposition Let $\abs{\cdot}$ be a valuation on $K$ such that property (3) holds with $c = 2$. Then $\abs{\cdot}$ satisfies the triangle inequality; i.e. $\abs{x + y} \le \abs{x} + \abs{y}$ for all $x, y\in K$.
Proof We are given that $\abs{x + y}\le 2\max(\abs{x}, \abs{y})$ for all $x, y\in K$. By induction we see that \[\abs{\sum_{i = 1}^{2^s} x_i}\le 2^s \max_{1\le i\le 2^s}\abs{x_i}\] for all positive integers $s$. In particular, if $2^{s - 1}\le n< 2^s$, \[ \abs{\sum_{i = 1}^{n} x_i}\le 2^s \max_{1\le i \le n}\abs{x_i} < 2n \max_{1\le i \le n}\abs{x_i} \] One immediate consequence is that $\abs{n} < 2n$ for all positive integers $n$. Now for all $x, y\in K$ and for all positive integers $n$, we have \[ \begin{align*} \abs{x + y}^n = \abs{(x + y)^n} &= \abs{\sum_{i = 0}^n \binom{n}{i} x_i y^{n - i}} \\ &\le 2(n + 1)\max_{0\le i\le n} \abs{\binom{n}{i} x^i y^{n - i}} \\ &\le 2(n + 1)\sum_{i = 0}^n \abs{\binom{n}{i} x^i y^{n - i}} \\ &\le 4(n + 1)\sum_{i = 0}^n \binom{n}{i} \abs{x^i y^{n - i}} \\ &= 4(n + 1)(\abs{x} + \abs{y})^n. \end{align*} \] Taking $n\to \infty$ gives $\abs{x + y}\le \abs{x} + \abs{y}$, as desired. $\boxed{}$
Corollary Let $\abs{\cdot}$ be a valuation on $K$. Then $\abs{\cdot}$ is equivalent to a valuation that satisfies the triangle inequality.
Proof Pick $c$ such that property (3) holds, and pick $r > 0$ such $c^r \le 2$. Then $\abs{\cdot}^r$ satisfies the triangle inequality. $\boxed{}$
Example / Exercise Let $K = \BQ$ and let $p$ be a rational prime. Let $v_p(x)$ be the exponent of the power of $p$ dividing $x$. Then the function given by $\abs{x} = p^{-v_p(x)}$ is a valuation on $\BQ$. (This is the standard $p$-adic valuation on $\BQ$, and is denoted $\abs{\cdot}_p$.)
Definition / Example The trivial valuation $\abs{\cdot}_0$ is the one that takes on the value of 1 on all of $K^\times$.
Proposition Suppose $\abs{\cdot}$ is a valuation on $K$, and let $r$ positive real number. Then $\abs{\cdot}^r$ is also a valuation on $K$.
Proof Obvious. $\boxed{}$
Definition Two nontrivial valuations $\abs{\cdot}_1$ and $\abs{\cdot}_2$ are equivalent if there exists a real number $r > 0$ such that $\abs{x}_1^r = \abs{x}_2$ for all $x\in K$.
Proposition Let $\abs{\cdot}$ be a valuation on $K$ such that property (3) holds with $c = 2$. Then $\abs{\cdot}$ satisfies the triangle inequality; i.e. $\abs{x + y} \le \abs{x} + \abs{y}$ for all $x, y\in K$.
Proof We are given that $\abs{x + y}\le 2\max(\abs{x}, \abs{y})$ for all $x, y\in K$. By induction we see that \[\abs{\sum_{i = 1}^{2^s} x_i}\le 2^s \max_{1\le i\le 2^s}\abs{x_i}\] for all positive integers $s$. In particular, if $2^{s - 1}\le n< 2^s$, \[ \abs{\sum_{i = 1}^{n} x_i}\le 2^s \max_{1\le i \le n}\abs{x_i} < 2n \max_{1\le i \le n}\abs{x_i} \] One immediate consequence is that $\abs{n} < 2n$ for all positive integers $n$. Now for all $x, y\in K$ and for all positive integers $n$, we have \[ \begin{align*} \abs{x + y}^n = \abs{(x + y)^n} &= \abs{\sum_{i = 0}^n \binom{n}{i} x_i y^{n - i}} \\ &\le 2(n + 1)\max_{0\le i\le n} \abs{\binom{n}{i} x^i y^{n - i}} \\ &\le 2(n + 1)\sum_{i = 0}^n \abs{\binom{n}{i} x^i y^{n - i}} \\ &\le 4(n + 1)\sum_{i = 0}^n \binom{n}{i} \abs{x^i y^{n - i}} \\ &= 4(n + 1)(\abs{x} + \abs{y})^n. \end{align*} \] Taking $n\to \infty$ gives $\abs{x + y}\le \abs{x} + \abs{y}$, as desired. $\boxed{}$
Corollary Let $\abs{\cdot}$ be a valuation on $K$. Then $\abs{\cdot}$ is equivalent to a valuation that satisfies the triangle inequality.
Proof Pick $c$ such that property (3) holds, and pick $r > 0$ such $c^r \le 2$. Then $\abs{\cdot}^r$ satisfies the triangle inequality. $\boxed{}$
Theorem ([Little] Ostrowski's Theorem) When $K = \BQ$, all nontrivial valuations on $K$ are equivalent to either $\abs{\cdot}_\infty$ or $\abs{\cdot}_p$ for some prime $p$.
Proof First suppose $\abs{n}\le 1$ for all positive integers $n$. Set $I = \{n\in \BZ: \abs{n} < 1\}$; note that $I$ is a proper ideal of $\BZ$. Since $\abs{\cdot}$ is nontrivial, $I$ is nonempty, so $I = (n)$ for some positive integer $n > 1$. However, if $\abs{n} < 1$, then $\abs{p} < 1$ for some prime $p$ dividing $n$, which means that $p\in (n)$. The only way this is possible is if $n = p$, or $I = (p)$. Thus $\abs{q} = 1$ for all primes $q\neq p$, which means that $\abs{n} = \abs{p}^{v_p(n)}$ for all $n\in \BQ^\times$. As $\abs{p} < 1$, this is equivalent to the $p$-adic valuation.
Now suppose there exists an integer $m$ such that $\abs{m} > 1$. For every positive integer $n > 1$, we have that $m^k < n^{1 + \fl{\frac{k\log m}{\log n}}}$. Thus if we write out $m^k$ in base $n$ and use the triangle inequality we have \[ \abs{m}^k = \abs{m^k}\le (n-1)\left(1 + \fl{\frac{k\log m}{\log n}}\right)\max(1, \abs{n}^{ \fl{\frac{k\log m}{\log n}}}) \] As $k\to \infty$, $\abs{m}^k$ grows larger than $1 + \fl{\frac{k\log m}{\log n}}$, so we must have $\abs{n} > 1$ for all positive integers $n > 1$. In addition, taking $k\to \infty$ shows that for any pair of positive integers $m, n > 1$, \[ \abs{m}\le \abs{n}^{\displaystyle\frac{\log m}{\log n}}\implies \frac{\log\abs{m}}{\log m} = \frac{\log\abs{n}}{\log n} \] Thus there is a real number $r$ such that $\abs{n} = n^r$ for all positive integers $n$. Thus for all $x\in \BQ, \abs{x} = \abs{x}_\infty^r$. Furthermore, $\abs{n} > 1$ for all positive integers $n$, so $r > 1$. Thus $\abs{\cdot}$ is equivalent to $\abs{\cdot}_\infty$. $\boxed{}$
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